Expert: Ahmed Salami - 11/1/2009

Question
I need help with this problem, please:

Find the center, vertices, and foci of the ellipse whose equation is given below.
(Show some work). Be sure to carefully label your answers.

9x2 + y2 + 10y + 16 = 0
Thank you!


Answer
Hi Cindy,
Re-arranging the equation by completing the square in y,
9x² + y² + 10y + 16 = 0
9x² + (y² + 10y + 25) - 25 + 16 = 0
9x² + (y² + 10y + 25) = 9
9x² + (y+5)² = 9
x² + (y+5)²/3² = 1
Now, comparing with the standard equation of an ellipse
X²/a² + Y²/b² = 1
with center (0,0), vertices; [(-a,0),(0,b),(a,0),(0,-b)] and foci; (-ea,0) and (ea,0)
where ea = √(a² - b²)

For x² + (y+5)²/3² = 1
a = 1 and b = 3
Notice that b > a, and so the foci lies on the y axis and not the x axis as usual. And the y values have been moved down by a value of 5 i.e y = Y-5
So, eb = √(b² - a²)
      = √(3² - 1²)
      = √8
      = 2√2

center = (0,0-5)
      = (0,-5)
vertices are (-1,0-5),(0,3-5),(1,0-5),(0,-3-5)
i.e (-1,-5),(0,-2),(1,-5),(0,-8)
foci are (0,-2√2 - 5) and (0, 2√2 - 5)


Regards